OCR MEI Further Pure Core AS 2020 November — Question 6 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2020
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeProperties of matrix operations
DifficultyModerate -0.3 This is a straightforward Further Maths question testing basic matrix properties: commutativity (simple multiplication check), determinant product rule (standard recall), and solving simultaneous equations from MN=I. All parts are routine applications of standard techniques with no novel insight required, making it slightly easier than average even for Further Maths.
Spec4.03b Matrix operations: addition, multiplication, scalar4.03h Determinant 2x2: calculation4.03m det(AB) = det(A)*det(B)4.03n Inverse 2x2 matrix

6 The matrices \(\mathbf { M }\) and \(\mathbf { N }\) are \(\left( \begin{array} { l l } \lambda & 2 \\ 2 & \lambda \end{array} \right)\) and \(\left( \begin{array} { c c } \mu & 1 \\ 1 & \mu \end{array} \right)\) respectively, where \(\lambda\) and \(\mu\) are constants.
  1. Investigate whether \(\mathbf { M }\) and \(\mathbf { N }\) are commutative under multiplication.
  2. You are now given that \(\mathbf { M N } = \mathbf { I }\).
    1. Write down a relationship between \(\operatorname { det } \mathbf { M }\) and \(\operatorname { det } \mathbf { N }\).
    2. Given that \(\lambda > 0\), find the exact values of \(\lambda\) and \(\mu\).
    3. Hence verify your answer to part (i).

Question 6:
AnswerMarks Guidance
6(a) λµ+2 λ+2µ
MN=
 
λ+2µ λµ+2
λµ+2 λ+2µ
NM =  
λ+2µ λµ+2
AnswerMarks
So M and N are commutativeM1
A1cao
AnswerMarks Guidance
[2]2.5
2.2aAttempt to show MN = NM
6(b) (i)
[1]2.2a oe eg det N = 1/det M
6(b) (ii)
⇒ λ = −2µ, −2µ2 + 2 = 1
AnswerMarks
⇒ µ = −1/√2 , λ = √2B1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1soi
elimination
AnswerMarks
or µ = −√2/2or λ = −1/ µ, −1/ µ +2 µ = 0
must be exact
AnswerMarks Guidance
6(b) (iii)
so det N = 1/det MB1
B1
AnswerMarks
[2]2.1
2.2a
Question 6:
6 | (a) | λµ+2 λ+2µ
MN=
 
λ+2µ λµ+2
λµ+2 λ+2µ
NM =  
λ+2µ λµ+2
So M and N are commutative | M1
A1cao
[2] | 2.5
2.2a | Attempt to show MN = NM
6 | (b) | (i) | det M × det N = 1 | B1
[1] | 2.2a | oe eg det N = 1/det M | ‘inverse’ B0
6 | (b) | (ii) | λµ + 2 = 1, λ + 2µ = 0
⇒ λ = −2µ, −2µ2 + 2 = 1
⇒ µ = −1/√2 , λ = √2 | B1
M1
A1
[3] | 3.1a
1.1
1.1 | soi
elimination
or µ = −√2/2 | or λ = −1/ µ, −1/ µ +2 µ = 0
must be exact
6 | (b) | (iii) | det M = λ2 – 4 = −2, det N = µ2 – 1 = − ½
so det N = 1/det M | B1
B1
[2] | 2.1
2.2a
6 The matrices $\mathbf { M }$ and $\mathbf { N }$ are $\left( \begin{array} { l l } \lambda & 2 \\ 2 & \lambda \end{array} \right)$ and $\left( \begin{array} { c c } \mu & 1 \\ 1 & \mu \end{array} \right)$ respectively, where $\lambda$ and $\mu$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Investigate whether $\mathbf { M }$ and $\mathbf { N }$ are commutative under multiplication.
\item You are now given that $\mathbf { M N } = \mathbf { I }$.
\begin{enumerate}[label=(\roman*)]
\item Write down a relationship between $\operatorname { det } \mathbf { M }$ and $\operatorname { det } \mathbf { N }$.
\item Given that $\lambda > 0$, find the exact values of $\lambda$ and $\mu$.
\item Hence verify your answer to part (i).
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2020 Q6 [8]}}